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jok3333 [9.3K]
3 years ago
7

Rhett is solving the quadratic equation 0= x2 – 2x – 3 using the quadratic formula. Which shows the correct substitution of the

values a, b, and c into the quadratic formula?

Mathematics
1 answer:
lesya692 [45]3 years ago
4 0

Answer:

The correct option is 1.

Step-by-step explanation:

If a quadratic equation is defined as

ax^2+bx+c=0

then the quadratic formula is

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

The given equation is

0=x^2-2x-3

Here, a=1, b=-2 and c=-3.

Substitute these values in the above formula.

x=\frac{-(-2)\pm \sqrt{(-2)^2-4(1)(-3)}}{2(1)}

x=\frac{2\pm \sqrt{(-2)^2-4(1)(-3)}}{2(1)}

Therefore, the correct option is 1.

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Well me myself and I would say yes!!!
8 0
3 years ago
What advice would you give someone who is about to graduate from college and become a teacher?
Kazeer [188]

Answer:

he /she must be patient as a teacher

6 0
2 years ago
It is known that diskettes produced by a cer- tain company will be defective with probability .01, independently of each other.
zheka24 [161]

Answer:

1.27%

Step-by-step explanation:

To solve this problem, we may consider a binomial distribution where a customer can either accept or reject (and return) the diskette package.

Lets consider  some aspects:

1. From the formulation of the exercise we know that a package is accepted if it has at most 1 defective diskette. So our event A is defined as:

A = 0 or 1 defective diskette

2. The probability of a diskette being defective is 0.01

3. Each package contains 10 diskettes.

If X is defined as number of defective diskettes in the package, the probability of X is given by a binomial distribution with probability 0.01 and n=10

X ~ Bin(p=0.01, n=10)

Let us remember the calculation of probability for the binomial distribution:

P(X=x)=nCx*p^{x}*(1-p)^{(n-x)} with x = 0, 1, 2, 3,…, n

Where

n: number of independent trials

p: success probability  

x: number of successes in n trials

In our case success means finding a defective diskette, therefore

n=10

p=0.01

And for x we just need 0 or 1 defective diskette to reject the package

Hence,

P(X=x)=10Cx*0.01^{x}*(1-0.01)^{(10-x)} with x = 0, 1

So,

P(A)=P(X=0)+P(X=1)

P(A)=10C0*0.01^{0}*(1-0.01)^{(10-0)} + 10C1*0.01^{1}*(1-0.01)^{(9)}

P(A)=0.99^{10}+10*0.01*0.99^{9}

P(A)=0.9957

Now, because we have 3 packages and we might reject just 1 of them, we can find this probability like this:

3*(1-P(A))*P(A)*P(A) = (1-0.9957)*0.9957*0.9957=0.0127

Finally, we have that the probability of returning exactly one of the three packages is 1.27%

3 0
3 years ago
Y= (x-1)^2 + 2 <br><br> change this equation into standard form
BaLLatris [955]

Answer:

x² - 2x + 3

Step-by-step explanation:

y = (x - 1)² + 2

= (x - 1)² + 2 [a - b]² = <em><u>a²- 2ab + b²</u></em>

= x²- 2x + 1 + 2

= x² - 2x + 3

Thus, the standard form of the equation is x² - 2x + 3

8 0
3 years ago
1. If x=3-√3,<br> prove that x²<br> +36<br> x²<br> = x²
valentina_108 [34]

Answer:

we get x^2=12-6\sqrt{3}

Step-by-step explanation:

We are given: x=3-\sqrt{3}

We need to find x^2

Note: Since question is not clear, I am assuming that we need to find x^2

Solving:

x=3-\sqrt{3} \\Taking \ square \ on \ both \ sides\\x^2=(3-\sqrt{3})^2\\

We know that (a-b)^2= a^2-2ab+b^2

Using formula and simplifying

x^2=(3)^2-2(3)(\sqrt{3})+(\sqrt{3})^2\\x^2=9-6\sqrt{3} +3\\x^2=12-6\sqrt{3} \\

So, we get x^2=12-6\sqrt{3}

3 0
2 years ago
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